Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int const N = 1e4 + 10;
using vp = vector<pair<int, int>>;
int x[N], y[N];
vp rx, ry;
inline int kth(int x, int k) { return (x >> k) & 1; }
inline int msb(int x) {
  for (int i = 1; i <= 16; x |= x >> i, i <<= 1)
    ;
  return (x + 1) >> 1;
}
void swap(int x[], int a, int b, vp &ret) {
  if (a == b) return;
  ret.push_back({a, b}), ret.push_back({b, a}), ret.push_back({a, b});
  swap(x[a], x[b]);
}
void exor(int x[], int a, int b, vp &ret) {
  ret.push_back({a, b});
  x[a] ^= x[b];
}
int to_base(int x[], int const n, vp &ret) {
  int r = 0;
  for (int c = 30, i; r < n && c >= 0; --c) {
    for (i = r; i < n && !kth(x[i], c); ++i)
      ;
    if (i >= n) continue;
    swap(x, i, r, ret);
    for (i = r + 1; i < n; ++i)
      if (kth(x[i], c)) exor(x, i, r, ret);
    ++r;
  }
  return r;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", x + i);
  for (int i = 0; i < n; ++i) scanf("%d", y + i);
  int rank_x = to_base(x, n, rx), rank_y = to_base(y, n, ry);
  if (rank_x < rank_y) return puts("-1"), 0;
  for (int i = 0, j = 0, k; i < rank_y; ++i, ++j) {
    for (; j < rank_x && msb(x[j]) > msb(y[i]); ++j) exor(x, j, j, rx);
    if (j >= rank_x || msb(x[j]) < msb(y[i])) return puts("-1"), 0;
    int cc = y[i] ^ x[j];
    for (k = j + 1; cc && k < rank_x; ++k) {
      if (msb(x[k]) == msb(cc)) {
        cc ^= x[k];
        exor(x, j, k, rx);
      }
    }
    if (cc) return puts("-1"), 0;
  }
  for (int i = 0, k = 0; i < rank_x; ++i)
    if (x[i]) swap(x, k++, i, rx);
  for (int i = rank_y; i < n; ++i)
    if (x[i]) exor(x, i, i, rx);
  reverse(begin(ry), end(ry));
  rx.insert(end(rx), begin(ry), end(ry));
  printf("%d\n", (int)rx.size());
  for (auto &p : rx) printf("%d %d\n", p.first + 1, p.second + 1);
  return 0;
}